We have an array of integers and we have to find two such elements in the array such that sum of these two elements is equal to the sum of rest of elements in array.

Examples:

Input : arr[] = {2, 11, 5, 1, 4, 7} Output : Elements are 4 and 11 Note that 4 + 11 = 2 + 5 + 1 + 7 Input : arr[] = {2, 4, 2, 1, 11, 15} Output : Elements do not exist

A **simple solution** is to consider every pair one by one, find its sum and compare the sum with sum of rest of the elements. If we find a pair whose sum is equal to rest of elements, we print the pair and return true. Time complexity of this solution is O(n^{3})

An **efficient solution** is to find sum of all array elements. Let this sum be “sum”. Now the task reduces to finding a pair with sum equals to sum/2.

Another optimization is, a pair can exist only if the sum of whole array is even because we are basically dividing it into two parts with equal sum.

**1-** Find the sum of whole array. Let this sum be “sum”

**2-** If sum is odd, return false.

**3-** Find a pair with sum equals to “sum/2” using hashing based method discussed here as method 2. If a pair is found, print it and return true.

**4-** If no pair exists, return false.

Below is C++ implementation of above steps.

`// C++ program to find whether two elements exist ` `// whose sum is equal to sum of rest of the elements. ` `#include<bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to check whether two elements exist ` `// whose sum is equal to sum of rest of the elements. ` `bool` `checkPair(` `int` `arr[],` `int` `n) ` `{ ` ` ` `// Find sum of whole array ` ` ` `int` `sum = 0; ` ` ` `for` `(` `int` `i=0; i<n; i++) ` ` ` `sum += arr[i]; ` ` ` ` ` `// If sum of array is not even than we can not ` ` ` `// divide it into two part ` ` ` `if` `(sum%2 != 0) ` ` ` `return` `false` `; ` ` ` ` ` `sum = sum/2; ` ` ` ` ` `// For each element arr[i], see if there is ` ` ` `// another element with vaalue sum - arr[i] ` ` ` `unordered_set<` `int` `> s; ` ` ` `for` `(` `int` `i=0; i<n; i++) ` ` ` `{ ` ` ` `int` `val = sum-arr[i]; ` ` ` ` ` `// If element exist than return the pair ` ` ` `if` `(s.find(val) != s.end()) ` ` ` `{ ` ` ` `printf` `(` `"Pair elements are %d and %d\n"` `, ` ` ` `arr[i], val); ` ` ` `return` `true` `; ` ` ` `} ` ` ` ` ` `s.insert(arr[i]); ` ` ` `} ` ` ` ` ` `return` `false` `; ` `} ` ` ` `// Driver program. ` `int` `main() ` `{ ` ` ` `int` `arr[] = {2, 11, 5, 1, 4, 7}; ` ` ` `int` `n = ` `sizeof` `(arr)/` `sizeof` `(arr[0]); ` ` ` `if` `(checkPair(arr, n) == ` `false` `) ` ` ` `printf` `(` `"No pair found"` `); ` ` ` `return` `0; ` `} ` |

Output:

Pair elements are 4 and 11

Time complexity : O(n). unordered_set is implemented using hashing. Time complexity hash search and insert is assumed as O(1) here.

This article is contributed by **Niteesh kumar**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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